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STRUCTURE OF FLEROVIUM ISOTOPES
By Prof. Lefteris Kaliambos (Natural Philosopher in New Energy). ( October 2014) Historically the discovery of the assumed uncharged neutron (1932) along with the invalid relativity (EXPERIMENTS REJECT RELATIVITY) led to the abandonment of the well-established electromagnetic laws, in favour of various contradicting nuclear theories, which could not lead to the nuclear structure. Under this physics crisis and using the charged UP and DOWN quarks , discovered by Gell-Mann and Zweig, I published my paper “ “Nuclear structure is governed by the fundamental laws of electromagnetism (2003), which led to my discovery of the new structure of protons and neutrons given by proton = + 5d + 4u = 288 quarks = mass of 1836.15 electrons neutron = + 4u + 8d = 288 quarks = mass of 1838.68 electrons The paper was also presented at a nuclear conference held at NCSR "Demokritos" (2002). Here one can see the 9 charged quarks in proton and the 12 ones in neutron able to give the charge distributions in nucleons for revealing the strong electromagnetic force for the nuclear binding in the correct nuclear structure by applying the laws of electromagnetism. You can see my papers of nuclear structure in my FUNDAMENTAL PHYSICS CONCEPTS . Note that according to my discovery of the LAW OF ENERGY AND MASS the mass defect in the nuclear structure is due to the photon mass of the emitting dipolic photon presented at the international conference "Frontiers of fundamental physics" (1993) organised by the natural philosophers M. Barone and F. Selleri , who gave me an award including a disc of the atomic philosopher Democritus. Nevertheless today many physicist continue to apply the wrong nuclear theories which lead to complications. Flerovium (Fl) is an artificial element, and thus a standard atomic mass cannot be given. Like all artificial elements, it has no stable isotopes. The first isotope to be synthesized was Fl-289 in 1999 (or possibly 1998). Flerovium has five confirmed isotopes, and possibly 2 nuclear isomers. The longest-lived isotope is Fl-289 with a half-life of 2.6 seconds. Note that the Fl-289 has 61 extra neutrons which fill all the 61 blank positions, but the pp repulsions of long range (large number) always overcome such pn bonds of short range leading to the decay. ' ' NUCLEAR STRUCTURE OF Fl-286, AND Fl-288 WITH S =0 It is well well-known that the structure of lead-164 (core) of high symmetry consists of 8 horizontal planes and 2 horizontal lines providing 44 blank positions for receiving extra neutrons with two bonds per neutron in order to construct the stable Pb-208. (See the fourth figure of lead at the bottom of the page).Similarly the structure of Fl-228 (core) with 114 protons and 114 neutrons (even number) consists of 8 horizontal planes of opposite spins, including four additional deuterons with S = +2 and S = -2 which exist over and under the structure of 8 horizontal planes, forming the up horizontal line (+UHL) and the down horizontal line (-DHL). So all these nucleons of the 8 horizontal planes and the +UHL and the -DHL give S = 0 . In general, the structure of Fl-228 (core) has S =0 and is similar to the structure of Pb-164, because the two additional vertical systems of p113n113 and p114n114 with S = 0 make symmetrical vertical rectangles. So the structure of the above unstable nuclides with even number of extra neutrons is based on the Fl-228 with S =0, because the total spin S =0 is due to the extra neutrons of opposite spins. For example the Fl-288 with S =0 has 60 extra neutrons with opposite spins. They fill 60 blank positions but the pp repulsions of long range (large number) always overcome such pn bonds of short range leading to the decay. ' ' 'NUCLEAR STRUCTURE OF Fl-285, AND Fl-289 ' After a careful analysis I found that the structure of such unstable nuclides, with odd number of extra neutrons, is based on another structure of the Fl-228 (core) having S = +2. In this case the one deuterons of the -DHL changes the spin from S = -1 to S = +1 giving S = +2, because it moves to +UHL for making horizontal bonds with a deuteron of the up horizontal line. For example the Fl-285 with S = +3/2 of 57 extra neutrons, has 28 extra neutrons of positive spins and 29 extra neutrons of negative spins . That is S = +2 + 28(+1/2) + 29(-1/2) = +3/2 Note that the 57 extra neutrons fill 57 blank positions, but the pp repulsions of long range (large number) always overcome such pn bonds of short range leading to the decay. Category:Fundamental physics concepts